$A=\left[\begin{array}{rr}22 & -19 & 37 & 61 \\11 & 22 & -85 &4 \\33 &16 &15 & 9 \\8 &9 &10 & -2\end{array}\right]$ $A_{3,1}=$
Solution: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{3,1}$ $A_{{3},{1}}$ is located on row ${3}$ of $A$ : $\left[\begin{array}{rr}22 & -19 & 37 & 61 \\11 & 22 & -85 &4 \\ {33} & {16} & {15} & {9} \\8 &9 &10 & -2\end{array}\right]$ $A_{{3},{1}}$ is also located on column ${1}$ of $A$. $\left[\begin{array}{rr}{22} & -19 & 37 & 61 \\{11} & 22 & -85 &4 \\ {\text{33}} & {16} & {15} & {9} \\{8} &9 &10 & -2\end{array}\right]$ Therefore, $A_{{3},{1}}={33}$. Summary $A_{3,1}=33$